If a + x = b + y = c + z + 1, where a, b, c, x, y, z are non-zero...

Given $x + a = y + b = z + c + 1$

Now \left|$\begin{cases}x & a + y & a + x \\ y & b + y & b + y \\ z & c + y & c + z\end{cases}$\right| = \left|$\begin{cases}x & a + y & a \\ y & b + y & b \\ z & c + y & c\end{cases}$\right|\left({C}_{3} \to {C}_{3} - {C}_{1}\right)$</p><p>$= \left|$\begin{cases}x & y & a \\ y & y & b \\ z & y & c\end{cases}$\right|\left({C}_{2} \to {C}_{2} - {C}_{3}\right)$</p><p>$= y\left|$\begin{cases}x & 1 & a \\ y & 1 & b \\ z & 1 & c\end{cases}$\right|$</p><p>${R}_{2} \to {R}_{2} - {R}_{1} and ${R}_{3} \to {R}_{3} - {R}_{1}$

y\left|\begin{cases}x & 1 & a \\ y - x & 0 & b - a \\ z - x & 0 & c - a\end{cases}\right| = y\left|\begin{cases}x & 1 & a \\ a - b & 0 & - ( a - b ) \\ z - x & 0 & c - a\end{cases}\right| = y ( a - b ) \left|\begin{cases}x & 1 & a \\ 1 & 0 & - 1 \\ z - x & 0 & c - a\end{cases}\right| = - y ( a - b ) ( c - a + z - x ) = y ( a - b )